If a stoichiometric ratio is taken, then
When sodium is burned in oxygen, Na2O2 peroxide is formed, also there is an admixture of Na2O (that is, at the beginning - peroxide, and as oxygen becomes smaller, the sodium residue reduces the peroxide to oxide).
However, one mole of any of the compounds contains two moles of sodium.
"\\nu(Na)=mass(Na)\/Molar(Na)=>\n\\nu(Na)=6,50g\/22,99=0,2827"
vsodium oxide="\\nu(Na)\/2=0,14135" moles
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